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Given the probability density function: $$ f(x) = Ce^x \quad 0<x<1 $$ To find the constant \(C\), we have to normalize the probability distribution. The normalization condition states that the integral of the probability density function over its entire range should equal 1: $$ \int_{0}^{1} f(x)dx = 1 $$ Substitute \(f(x)\) with the given function: $$ \int_{0}^{1} Ce^x dx = 1 $$ Next, integrate the function with respect to \(x\): $$ C\int_{0}^{1} e^x dx = 1 $$ We have: $$ C[e^x]_{0}^{1} = 1 $$ Plugging in the limits, we get: $$ C(e^1 - e^0) = 1 $$ Since \(e^0 = 1\), the equation becomes: $$ C(e - 1) =1 $$ Solving for \(C\), we get: $$ C = \frac{1}{e - 1} $$

Now that we have found the constant \(C\), our probability density function becomes; $$ f(x) = \frac{1}{e - 1} e^x \, 0<x<1 $$ To simulate such a random variable, we can use the Inverse Transform Sampling technique. This method involves the following steps: 1. Compute the cumulative distribution function (CDF) for the random variable X. 2. Generate a random number \(u\) uniformly distributed between 0 and 1. 3. Find the inverse of the CDF (denoted \(F^{-1}(u)\)). 4. The random variate \(x = F^{-1}(u)\) will follow the given probability density function \(f(x)\). Let's find the cumulative distribution function (CDF) F(x): $$ F(x) = \int_{0}^{x} \frac{1}{e-1} e^t dt $$ Integrate the function and apply the limit: $$ F(x) = \frac{1}{e-1} [e^t]_{0}^{x} = \frac{1}{e-1}(e^x - 1) $$ To find the inverse of the CDF, we solve for \(x\) in terms of \(F^{-1}(u)\): $$ u = \frac{1}{e-1}(e^x - 1) $$ Solving for \(x\): $$ x = \ln(u(e - 1) + 1) $$ Now we have the inverse of the CDF. By generating random numbers \(u\) between 0 and 1 and computing the corresponding \(x\) values as per the inverse CDF formula, the generated \(x\) values will follow the given probability density function \(f(x)\).